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Chapter 6: Problem 55

Solving a First-Order Linear Differential Equation In Exercises \(51-58\) , solve the first-order differential equation by any appropriate method. $$ \left(2 y-e^{x}\right) d x+x d y=0 $$

Short Answer

Expert verified
The solution of the given first-order linear differential equation is \( y= \frac{e^{x}(x-1) + C}{x^2}\).

Step by step solution

01

Writing the equation in differential form

Rearranging the terms to separate the differentials gives: \(dy = -\frac{(2y-e^x)}{x} dx \)
02

Rewriting Equation in Standard Form

The standard form of a first-order differential equation is \(dy/dx + P(x)y = Q(x)\). The given equation can be rewritten as: \( dy/dx + \frac{2}{x} y = \frac{e^{x}}{x} \). Here, \( P(x) = \frac{2}{x} \) and \( Q(x) = \frac{e^{x}}{x}\).
03

Find the Integrating Factor

The integrating factor is \( e^{\int P(x) dx} \). Therefore, the integrating factor is \( e^{\int 2/x dx} = x^2 \).
04

Multiply through by the integrating factor

Multiplying every term of the equation by \( x^2 \) gives: \( x^2 dy/dx + 2x y = x e^{x} \).
05

Find the solution

The left side is the derivative of \( yx^2 \) with respect to \( x \). So integrate the equation, yielding: \( \int (yx^2)' dx = \int x e^{x} dx \). Use integration by parts on the right hand side, where \( u=x \), \( dv=e^{x} dx \), \( du=dx \), and \( v=e^{x} \). So the integral of \(x \cdot e^{x} dx\) is equal to \(x \cdot e^{x} - e^{x}\). Therefore, the solution is \( yx^2= x e^{x} - e^{x} + C \). Solve for y to get the final solution: \( y= \frac{e^{x}(x-1) + C}{x^2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Factor
When faced with a first-order linear differential equation, finding an integrating factor simplifies the equation and is key to solving it. An integrating factor is a function, often denoted as \( \mu(x) \), that makes the left side of the equation into the derivative of a product of functions.

In our example, the equation \( \frac{dy}{dx} + \frac{2}{x}y = \frac{e^x}{x} \) can be adapted using an integrating factor. We determine this factor by calculating \( e^{\int P(x) \, dx} \), where \( P(x) = \frac{2}{x} \).

  • Compute the integral \( \int \frac{2}{x} \ dx = 2 \ln |x| \)
  • Raise \( e \) to this power, resulting in the integrating factor \( x^2 \)
Using the integrating factor, multiply throughout the equation to transform it, substantiating the equation's left side into a perfect derivative.
Differential Form
Before employing any methods to solve differential equations, it's crucial to manipulate them into a differential form. This approach makes it easier to manage and apply further solution techniques. The differential form is commonly represented as \( M(x, y) \, dx + N(x, y) \, dy = 0 \).

In the original exercise, the provided equation was \((2y - e^x) \, dx + x \, dy = 0\).

  • It denotes a relationship where components are expressed with respect to differentials \( dx \) and \( dy \).
  • Rearranging terms enabled separating the differentials into a standard form \( dy = -\frac{(2y-e^x)}{x} \ dx \).
Integration by Parts
Integration by parts is a versatile technique for solving integrals. It can transform complex integrals into simpler ones by integrating the product of two functions. It is based on the formula: \[\int u \, dv = uv - \int v \, du\]

This technique was pivotal in the solution's final step. For \( \int x e^x \, dx \), we selected:

  • \( u = x \), \( dv = e^x \, dx \)
  • Then \( du = dx \), and \( v = e^x \)
Applying the formula gives us: \( x e^x - \int e^x \, dx = x e^x - e^x \). This calculation helped solve for the function \( y \).
Standard Form of Differential Equations
The standard form of a first-order linear differential equation is crucial for applying systematic solution methods. It is represented as \( \frac{dy}{dx} + P(x)y = Q(x) \). This organization aligns the equation for further operations, like locating an integrating factor or solving using specific methods.

In our case, the standard form derived was:
\[ \frac{dy}{dx} + \frac{2}{x} y = \frac{e^x}{x} \]

  • This format highlights that \( P(x) = \frac{2}{x} \)
  • And \( Q(x) = \frac{e^x}{x} \)
The transformation to this form is an essential pre-step in solving the differential equation and ensures consistency in applying methods like integrating factors.

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Most popular questions from this chapter

Finding a Particular Solution In Exercises \(25-28\) , find an equation of the graph that passes through the point and has the given slope. $$ (8,2), \quad y^{\prime}=\frac{2 y}{3 x} $$

Determining if a Function Is Homogeneous In Exercises \(67-74,\) determine whether the function is homogeneous, and if it is, determine its degree. A function \(f(x, y)\) is homogeneous of degree \(n\) if \(f(t x, t y)=t^{n} f(x, y)\) $$ f(x, y)=\frac{x^{2} y^{2}}{\sqrt{x^{2}+y^{2}}} $$

Solving a Homogeneous Differential Equation In Exercises \(75-80\) , solve the homogeneous differential equation in terms of \(x\) and \(y .\) A homogeneous differential equation is an equation of the form \(M(x, y) d x+N(x, y) d y=0,\) where \(M\) and \(N\) are homogeneous functions of the same degree. To solve an equation of this form by the method of separation of variables, use the substitutions \(y=v x\) and \(d y=x d v+v d x\) $$ \left(x^{3}+y^{3}\right) d x-x y^{2} d y=0 $$

Weight Gain A calf that weighs \(w_{0}\) pounds at birth gains weight at the rate \(d w / d t=1200-w,\) where \(w\) is weight in pounds and \(t\) is time in years. Solve the differential equation.

Solving a Bernoulli Differential Equation In Exercises \(59-66,\) solve the Bernoulli differential equation. The Berchoulli equation is a well-known nonlinear equation of the form $$ y^{\prime}+P(x) y=Q(x) y^{n} $$ that can be reduced to a linear form by a substitution. The general solution of a Bernoulli equation is $$ y^{1-n} e^{\int(1-n) P(x) d x}=\int(1-n) Q(x) e^{f(1-n) P(x) d x} d x+C $$ $$ y y^{\prime}-2 y^{2}=e^{x} $$
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